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When a resistance of `100Omega` is connected in series with a galvanometeer of resistance R, its range is V. To double its range, a resistance of `1000Omega` is connected in series. Find R.A. `700Omega`B. `800Omega`C. `900Omega`D. `100Omega` |
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Answer» Correct Answer - C When a resistance of `100 Omega` is connected in series current, `i=(V)/(100+R)` ….(i) When a resistance of `100 Omega` is connected in series, the its range double current, `i=(2V)/(1100+R)` …(ii) From Eqs. (i) and (ii), `(V)/(100+R)=(2V)/(1100+R)` `rArr " " R = 900 Omega` |
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